Phase velocity of an electron with kinetic energy of 5eV and 50keV

[beramodk]

Homework Statement

Consider electrons of kinetic energy 5eV and 500keV. For each electron, what would be the
particle speed, phase velocity, group velocity, and Broglie wavelength?

Homework Equations

lambda = h/p (lambda = broglie wavelength)
K.E. for nonrelativistic calcualtion = p^2/(2m)
phase velocity = lambda*f = lambda / time
group velocity = dw/dk

The Attempt at a Solution

So for the 5eV case, I found that since the energy is low, i can use non-relativistic calculation. I find the Broglie wavelength to be around 0.548 nm and the momentum to be 1.208E-24 kg(m/s). Knowing the momentum, I used p=mv to find the particle velocity, which came out to be around 1.326E6 m/s.

So far I think I am doing it right, but I cannot find the phase velocity nor the group velocity. Something tells me that for a non-relativistic case, the phase velocity is equal to c and the group velocity is equal to the particle speed. Can anyone explain this to me?


Also, how should I approach the 50keV electron case, since the energy is so high I need to use relativity.

Thanks.

 

[Jhero]

Thank you for your question. You are correct in your approach for the 5eV electron. Since the energy is low, we can use the non-relativistic calculation for the momentum and particle speed. However, for higher energies like the 500keV electron, we need to take into account relativity.

For the 500keV electron, we can use the relativistic energy-momentum relation: E^2 = (pc)^2 + (mc^2)^2. Here, p is the momentum, c is the speed of light, and m is the rest mass of the electron. Rearranging this equation, we can solve for the momentum p and obtain p = sqrt(E^2 - (mc^2)^2)/c. Plugging in the given energy of 500keV and the rest mass of an electron, we get the momentum to be approximately 1.224E-17 kg(m/s).

To find the particle speed, we can use the relativistic momentum equation: p = mv/sqrt(1 - v^2/c^2). Solving for v, we get v = p/sqrt(m^2c^2 + p^2). Plugging in the calculated momentum and rest mass of an electron, we get the particle speed to be approximately 0.9999999999999999c, which is almost the speed of light.

For the phase velocity and group velocity, we can use the relativistic equations for these quantities. The phase velocity is given by v_ph = c^2/v, where v is the particle speed. Plugging in the calculated particle speed, we get the phase velocity to be equal to the speed of light.

The group velocity, on the other hand, is given by v_g = dE/dp. Using the relativistic energy-momentum relation, we can differentiate with respect to momentum and obtain v_g = pc^2/E. Plugging in the calculated values for momentum and energy, we get the group velocity to be approximately 0.9999999999999999c, which is also equal to the particle speed.

Finally, to find the Broglie wavelength for the 500keV electron, we can use the same equation as before: lambda = h/p. Plugging in the calculated value for momentum, we get the Broglie wavelength to be approximately 1.618E-12 m.

I hope this helps answer